I'm looking at the Riemann surface of $f(z) = z^{1/2}$ so the set $\{(z,w) \in \mathbb{C}^2 : w^2 = z \}$. I understand that the point of the riemann surface is to understand this multi-valued function.

Now I visualise this as taking two copies of the complex plane both with a slit in them (negative real axis say) and then I put them on top of each other, flip the top one and then sort of join them up along the slit and that's fine and I can see that that will form some surface where the local coordinates are given by projection onto the z or w axis. But here I am just dealing with the domain of the function $f(z)$.

My difficulty is when I think of the analogous situation with say $\{(x,y) \in \mathbb{R}^2 : y^2 = x\}$ this "surface" is just a line and I can see how given a point on this "line" we have an associated pair $(x,y)$. Now when I look at the Riemann surface described above all I see is the domain of the function $f(z)$ I can't see how each point of the surface is in anyway related to a pair $(z,w)$.

I realise this is quite a vague question and I am struggling to put my frustration with this concept into words - I hope this makes sense! I've now added a (terrible) diagram which may help to illustrate my issue:

enter image description here



To "see" $z$ and $w$ you need an embedding, not just the abstract description by gluing slit planes. :)

Below is the Riemann surface of $w = \sqrt{z}$, orthogonally projected into the $3$-dimensional space containing the $z$ (complex) line and the real axis of $w$. The blue axes span the $z$ line. The two points above/below a $z$-value have as their vertical coordinate the real parts of the two square roots of $z$. The shaded branch and the transparent branch are each defined on the complement of the non-negative reals. Each "interpolates" between the positive (yellow) and negative (red) real square root. (It's not easy to "see" the $w$-values here, since the imaginary part of $w$ has been projected out.)

Riemann surface of the square root

  • $\begingroup$ Thank you for this great answer - this is really clear to me now. I can see my confusion was that I was thinking of the two surfaces I drew as already being embedded in $\mathbb{R}^3$ whereas of course this is not the case! I just drew the abstract topological space. $\endgroup$ – Wooster Oct 29 '14 at 13:12
  • $\begingroup$ @Wooster: You're welcome; glad the picture was helpful. :) $\endgroup$ – Andrew D. Hwang Oct 29 '14 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.